T3 Property is Hereditary

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Theorem

Let $T = \left({S, \tau}\right)$ be a topological space which is a $T_3$ space.

Let $T_H = \left({H, \tau_H}\right)$, where $\varnothing \subset H \subseteq S$, be a subspace of $T$.


Then $T_H$ is a $T_3$ space.


Proof

Let $T = \left({S, \tau}\right)$ be a $T_3$ space.

Then:

$\forall F \subseteq S: \complement_S \left({F}\right) \in \tau, y \in \complement_S \left({F}\right): \exists U, V \in \tau: F \subseteq U, y \in V: U \cap V = \varnothing$

That is, for any closed set $F \subseteq S$ and any point $y \in S$ such that $y \notin F$ there exist disjoint open sets $U, V \in \tau$ such that $F \subseteq U$, $y \in V$.


We have that the set $\tau_H$ is defined as:

$\tau_H := \left\{{U \cap H: U \in \tau}\right\}$


Let $F \subseteq H$ such that $F$ is closed in $H$.

Let $y \in H$ such that $y \notin F$.

From Closed Set in Topological Subspace $F$ is also closed in $T$.

Because $T$ is a $T_3$ space, we have that:

$\exists U, V \in \tau: F \subseteq U, y \in V, U \cap V = \varnothing$

As $F \subseteq H$ and $y \in H, y \notin F$ we have that:

$F \subseteq U \cap H, y \in V \cap H: \left({U \cap H}\right) \cap \left({V \cap H}\right) = \varnothing$

and so the $T_3$ axiom is satisfied in $H$.

$\blacksquare$


Sources